The tide can be modelled using simple harmonic motion - HSC - SSCE Mathematics Extension 1 - Question 13 - 2016 - Paper 1

The tide is increasing at the fastest rate when the derivative with respect to time is maximized. We differentiate the function:

$\frac{dx}{dt} = -4 \cdot \left(-\frac{4\pi}{25}\sin\left(\frac{4\pi}{25} t\right)\right) = \frac{16\pi}{25}\sin\left(\frac{4\pi}{25} t\right)$

Setting this derivative to zero gives:

$\sin\left(\frac{4\pi}{25} t\right) = 0$

The sine function is zero at integer multiples of $\\pi$, therefore:

$\frac{4\pi}{25} t = n\pi$

Solving for $t$: $t = \frac{25n}{4}$ for integers n. The fastest increase occurs at the first positive solution after 2 am, which will be at $t = 6.25$ hours or 8:15 am.

The height of the projectile at time $t$ is given by $y = u \sin \theta \cdot t - \frac{1}{2} g t^{2}$. Setting $g = 10$, we have:

$y = u \sin \theta \cdot t - 5t^{2}$

To find the maximum height, set the derivative to zero:

$\frac{dy}{dt} = u \sin \theta - 10t = 0$

This gives:

$t = \frac{u \sin \theta}{10}$

Substituting back into the height equation provides:

$y_{max} = u \sin \theta \cdot \frac{u \sin \theta}{10} - 5(\frac{u \sin \theta}{10})^{2}$

Solving leads to:

$y_{max} = \frac{u^{2} \sin^{2} \theta}{20}$

The time to reach the wall horizontally is determined by:

$x = u \cos \theta \cdot t$

Assuming the wall is at a distance of $x = \frac{30}{\cos(30^{\circ})}$ horizontally. Using $\theta = 30^{\circ}$, we calculate:

$t = \frac{x}{u \cos(30^{\circ})}$

This time can be substituted back to find the vertical position using:

$y = 20 + u \sin(30^{\circ})t - 5t^{2}$

By solving, we find:

$y = \frac{125}{4}$

Using the rebound velocity of 10 m/s horizontally, we analyze the vertical motion after rebounding from a height of $\frac{125}{4}$ m:

Using the equation:

$y = h_{initial} - \frac{1}{2} g t^2$

Solving for $t$ yields:

$\frac{125}{4} = \frac{1}{2} \cdot 10 t^{2}\Rightarrow t^{2} = \frac{125}{20} \Rightarrow t = \sqrt{6.25} = 2.5 s$

Using the horizontal speed of 10 m/s after rebounding, the distance traveled while falling for 2.5 seconds is:

$d = v \cdot t = 10 imes 2.5 = 25 m$

To prove that CMDE is a cyclic quadrilateral, we need to show that angles C and E sum up to 180 degrees. Since the angle subtended on the circumference by the same arc (CE) must equal each other.

Thus: $\angle C + \angle E = 180^{\circ}$

Hence, CMDE is cyclic.

Since CMDE is cyclic, by the properties of cyclic quadrilaterals, we have: $\angle ACF = \angle ADE$.

Also, since 'MF' intersects 'AB' at 'F', and both angles are equal, we conclude that: $MF \perp AB$.